Name the following triangle :
$∠ A = 60^{o}, ∠ B = 90^{o}, ∠ C = 30^{o}$

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Solution

Since $∠ B = 90^{\circ}$ ,Therefore it is an right angle triangle right angled at B.

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Similar questions

△ A B C, ∠ A = 90^{o}, ∠ B = 60^{o}, ∠ C = 30^{o}

△ A B C, ∠ A = 90^{o}

∠ B = 30^{o}

△ D E F, ∠ E = 30^{o}

∠ F = 60^{o}

Δ A B C, ∠ B = 90^{o}, ∠ C = 60^{o}, ∠ A = 30^{o}, A C = 16 c m

B C

c m

\frac{{tan}^{2} 60^{o} + 4 {sin}^{2} 45^{o} + {sin}^{2} 90^{o}}{3 {sec}^{2} 30^{o} + {csc}^{2} 60^{o} - {cot}^{2} 30^{o}}

\frac{{tan}^{2} 60^{o} + 4 {cos}^{2} 45^{o} + 3 {sec}^{2} 30^{o} + 5 {cos}^{2} 90^{o}}{c o sec 30^{o} + sec 60^{o} - {cot}^{2} 30^{o}}

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